Question

If f(x) = |(x2 − 4)(x2 + 2)|, how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?

Answer 1

@Astrophysics @ganeshie8 @Michele_Laino

Answer 2

help..

Answer 3

I'm pondering...

Answer 4

Oh, sorry

Answer 5

we have to note that we have the subsequent inequality:
\[\Large {x^2} - 4 < 0,\quad {\text{if }} - 2 < x < 2\]

Answer 6

am I right?

Answer 7

honestly, im confused

Answer 8

why?

Answer 9

it is a simple inequality

Answer 10

no i understand the inequality, im confused by how you got it

Answer 11

ok! I wrote that inequality, because inside the interval [0,1] your function can be rewritten as follows:
\[\Large f\left( x \right) = - \left( {{x^2} - 4} \right)\left( {{x^2} + 2} \right) = \left( {4 - {x^2}} \right)\left( {{x^2} + 2} \right)\]

Answer 12

otherwise any other would be out of bounds

Answer 13

no, it is a step in order to get rid the absolute value symbol

Answer 14

oh, okay.

Answer 15

now, we can apply the theorem af mean value

Answer 16

so, we have to compute this:
\[\Large \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}} = \frac{{3 \times 3 - 4 \times 2}}{1} = ...\]

Answer 17

\[f'(c) = \frac{ f(b) - f(a) }{ b - a }\]

Answer 18

1

Answer 19

are you sure?

Answer 20

\[\large \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}} = \frac{{3 \times 3 - 4 \times 2}}{1} = \frac{{6 - 8}}{1} = ...?\]

Answer 21

isnt 3x3 9?

Answer 22

sorry I have made a big error, you are right

Answer 23

it's okay :)

Answer 24

sorry again!!

Answer 25

no apologies necessary! thank you for your help

Answer 26

next we have to compute the first derivative at a generic point x, namely we have to compute the first derivative of this function:
\[\Large f\left( x \right) = \left( {4 - {x^2}} \right)\left( {{x^2} + 2} \right)\]

Answer 27

what do you get?

Answer 28

\[-4x(x^2-1)\]

Answer 29

correct!

Answer 30

then we have to solve this equation:
\[ \Large - 4x\left( {{x^2} - 1} \right) = 1\]

Answer 31

it is not a simple equation

Answer 32

\[x^3 = -\frac{ 1 }{ 2 }\] ?

Answer 33

how did you do to write that expression?

Answer 34

\[-4x(x^2-1) = 1\]\[\frac{ 1 }{ -4x } = x^2\] \[\frac{ 1 }{ -4x } +1 = x^2\] \[1+1 = -4x^3\] \[- \frac{ 1 }{ 2 } = x^3\]

Answer 35

I don't understand this step:
\[\frac{1}{{ - 4x}} = {x^2}\]

Answer 36

anyway it is not a solution of our equation

Answer 37

Sorry, i forgot the -1

Answer 38

How would you get the proper solution?

Answer 39

I questioned wolfram alpha, here are the solutions:

Answer 40

So, 3 solutions in the interval

Answer 41

we have two solutions, inside the interval [0,1]

Answer 42

Whoops, my bad! I looked outside the interval

Answer 43

do you see them?

Answer 44

Yes. Since one of them is in the negatives, it is not in the interval and does not matter to answer the question

Answer 45

yes! that's right!

Answer 46

Thank you!

Answer 47

:)

Answer 48

I tried doing the same thing on this problem, but I keep getting no solutions:
A particle moves along the x-axis with position function s(t) = xe^x. How many times in the interval [−5, 5] is the velocity equal to 0?

Answer 49

velocity = (s(t))'
take derivative and let it =0 to solve for x .